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Since the topology is discrete, each subset of $X$ is open and the Borel $\sigma$-algebra is the collection of subsets of $X$.

We have to use Halmos' definition, with Dudley one the two $\sigma$-algebras coincide. The compact subsets of $X$ are finite (and $G_{\delta}$ since they are open), hence the smallest $\sigma$-algebra containing them contains all the countable subsets of $X$, and their complement. Since
$$\{A\subset X, A\mbox{ or }X\setminus A\mbox{ is at most countable}\}$$
is a $\sigma$-algebra, it's actually the Baire $\sigma$-algebra of $X$.

We can also use @t.b. argument: to see that $|X\times X|=|X|$, apply Zorn's lemma to $$P:=\{(A,g), A\subset X, f\colon A\times A\to A\mbox{ is a bijection}\},$$
with partial order $(A_1,f_1)\leq (A_2,f_2)$ if and only if $A_1\subset A_2$ and $g_{\mid A_1\times A_1}=f$. It shows that $(X,f)$ is maximal for some $f$. Then take $x_0\in X$, $S:=\{x_0\}\times X$, which is uncountable, with uncountable complement. Then $f(\{x_0\}\times X)$ does the job.

Can you give more details? (I assumed the large cardinality to avoid these problem, but it's probably not needed).
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Davide GiraudoAug 4 '12 at 17:31

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Davide, if $X$ is uncountable it has a subset of size $\aleph_1$; we can partition this set into two uncountable sets (in fact to $\aleph_1$ uncountable sets). Add the rest of $X$ to one of these, then you have an uncountable set whose complement is also uncountable.
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Asaf KaragilaAug 4 '12 at 17:33

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You have $\# X = \#(X \times X)$ for every infinite set. If $X$ is uncountable then every slice $\{x_0\} \times X$ is uncountable and has uncountable complement.
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t.b.Aug 4 '12 at 17:36

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Here's a nice answer by Andres Caicedo giving a detailed explanation of Asaf's suggestion in his second comment. (It's always dangerous to answer these measure theoretic questions with all those set theorists around...)
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t.b.Aug 4 '12 at 17:47